[RESOLVED] Code doubles MYSQL ENTRIES

[glima]

Guys can you please help me with this piece of code?
I dont undestand why it runs the SQL twice :

     foreach ($_POST as $key => $value){
     $$key = $_POST[$key];

 if(preg_match($pattmusic , $key, $matches, PREG_OFFSET_CAPTURE)){
      if(preg_match($pattmusicStyle , $key, $matches, PREG_OFFSET_CAPTURE)){
  $style = $value;
  }elseif(preg_match($pattmusicSinger, $key, $matches, PREG_OFFSET_CAPTURE)){		
  $singer= $value;
  if(!empty($style ) && !empty($singer)) {
  $insert_qr = "INSERT INTO album (albumID, singer, style) VALUES( '".$singer."', '".$style."')";}
  }
      mysql_select_db($database_dbConnect, $dbConnect);
  mysql_query($insert_qr, $dbConnect) or die(mysql_error());

I have 3 patterns:

$pattmusic = '/music/';
$pattmusicStyle = '/music_Style/';

$pattmusicSinger = '/music_Singer_/';

to enter the condition it first checks if it is from the MUSIC group... then it divides in two subcategories... music style and music instrument.
and in the last conditional it assigns the value to the query variable to be executed.. but its being executed and inserting the same record twice.
In this form... I have pairs of fields that has to be inserted both at the same time... because the user can add more field pairs on the fly with javascript..:

Music Style Music Instrument
{field 1} {field 1}
{field 2} {field 2}

Thanks for you help 🙂

[laserlight]

You might want to indent the code so you can see the flow of control better:

foreach ($_POST as $key => $value){
    $$key = $_POST[$key];

if(preg_match($pattmusic , $key, $matches, PREG_OFFSET_CAPTURE)){
    if(preg_match($pattmusicStyle , $key, $matches, PREG_OFFSET_CAPTURE)){
        $style = $value;
    }elseif(preg_match($pattmusicSinger, $key, $matches, PREG_OFFSET_CAPTURE)){
        $singer= $value;
        if(!empty($style ) && !empty($singer)) {
            $insert_qr = "INSERT INTO album (albumID, singer, style) VALUES( '".$singer."', '".$style."')";
        }
    }
    mysql_select_db($database_dbConnect, $dbConnect);
    mysql_query($insert_qr, $dbConnect) or die(mysql_error());

Clearly, the query will be executed for as many times as the loop runs with the first if condition being true (assuming that an SQL statement error does not kill the script before the last such iteration).